Poll: A little math problem
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Cheeze, you are not convincing me of anything by twisting my words to make it seem like I am saying things I did not say, or trying to reinvent the meaning of my words by scrambling the references. The fact you are employing such coy debating tactics says to me that you really don't have an argument left and are just arguing to argue.Cheeze_Pavilion post=18.73797.842068 said:
Okay, then. Let's start over from the beginning. We are given that the shopkeeper has two beagles, but she doesn't know their genders. She asks the person bathing them if there is at least one male. The person bathing them answers "yes". So, we know that at least one puppy is a male, but we don't know which one. So, we start by saying that maybe he meant that dog1 is the male puppy?I use the labels dog1 and dog2 to allow me to mentally track which dog I am referencing. Then, we see that dog2 is equally likely to be a male or a female. But, then again, maybe the person bathing the puppies is really talking about dog2. In that case, we have dog1 that is the one equally likely to be male or female. Now, I could make guesses about other possible processes that may have occurred that restrict the probabilities, but I can not reliably know that those processes have occurred. We are forced to assume those processes either have not occurred or do not have an effect because we have no means of gathering any information about them. Further, because we don't know which puppy is being referenced with the "yes" answer from the bather, we must include all possible relevant combinations of references that the bather could be making to the particular puppies in question(please don't twist my logic with crap like he could have been referencing a puppy out the window, down the street, and into the next state). With these assumptions, we obtain three unique possible outcomes(the key word being unique)Cheeze_Pavilion post=18.73797.842125 said:
dog1 dog2
M F
F M
M M
Among these three cases, there is only one case in which one is male and the other one is also male, irrespective of which "one" and "other" refer. So, we obtain a probability of 1/3 that the other one is also male.
Now, why do we not consider other potential influences on the probabilities, such as breeder selection. I have been proposing that the reason is that we have no means of obtaining information regarding the breeder's processes or how those processes affect the probabilities. For all we know, the breeder could select the puppies such there is never two males in a pair. But then this is unverifiable speculation. Thus, we can't really use it to answer the problem.
Well, more correctly, they're ones where the puppy-washer didn't need to look at the other pair. Lukeje's notation kinda obscures the fact that it doesn't matter whether or not the puppy-washer did look at the other one.Cheeze_Pavilion post=18.73797.842103 said:
And there are those other ones, without "(...)," which denote cases where you have to look at the other one.
Now, if you rewrite the Three-Card Problem to say that someone looks at the card and then tells you whether or not the card includes some red, and she says "Yes, there is a red side," the chance that the other side is red is only 50%. Which is what you've got here. Except there are two white-red cards for every red-red card, so you end up with 33%.
-- Alex
There are two cases, but they overlap on the unique dog1=M/dog2=M configuration. Maybe if we drew this as a Venn-diagram of unique configurations. The set of "dog1 is male" contains the elements dog1=M/dog2=F and dog1=M/dog2=M. The set of "dog2 is male" contains the elements dog1=F/dog2=M and dog1=M/dog2=M. The two sets overlap at the element dog1=M/dog2=M. So, despite the dog1=M/dog2=M configuration occurring in each of the cases above, we can only count it once. Does that seem reasonable?Cheeze_Pavilion post=18.73797.842257 said:
EDIT: Saw your edit after posting. Okay, fair enough. Let me try to be more precise with my use of words. We have two different cases, dog1 as male and dog2 as male. Among these two cases are three unique permutations. The three unique permutations are what I am using to calculate the probability because I don't know which of the two cases has actually occurred.
EDIT AGAIN: Changed the M/M to be dog1=M/dog2=M to be more clear that I am talking about the same element.
But if both are male, why would he be talking about Dog 2 as the male puppy. You have to understand that the order isn't established beforehand in some arbitrary fashion, it's established by which he checks first. Whichever is checked first is the first dog, whichever is checked second is the second. Therefore, if both are male, Dog 1 is always the "yes" male, and the other dog is Dog 2 (this doesn't actually matter, but may be helpful in understanding it).Cheeze_Pavilion post=18.73797.842392 said:
...Cheeze_Pavilion post=18.73797.842413 said:
Again, that doesn't actually matter. Whichever way it goes, he examines Dog 1 first, so Dog 1's gender is already known when you get to Dog 2. So if Dog 1's gender is male, it falls in these two situations:
MM
MF
Only if Dog 1 is female is there another case:
FM
Now, seriously. Is there really much left to argue about here?
This problem isn't really as complex as a lot of these posts make it sound. Someone brought up stuff about breeder selection, and that we don't know how he knows one of the dogs is male. But unless we assume something illogical like that some unknown circumstance might make both of them male, it doesn't even matter how he knows this. All that matters is that we know one or more of the dogs is male; where this information came from is irrelevant. There are only 2 steps here:
1. One or more of the dogs is male, so the possibilities are M/M, M/F, and F/M. Presumably, these are all equally probable.
2. The "Other" dog is best described as the dog not chosen from the pair to prove that one of them is male, so in each set the gender of this dog is M, F, and F, respectively. And that's it. 1/3 chance of being male.
The source of the dogs does not matter, and neither does anything else that could affect the dogs' genders, simply because we don't know any of this. It's just as likely that they could only be both male for some reason as it is that one of them must be female for some reason. If we had any information like that, we would already know the answer to 100% accuracy, and this wouldn't even be a problem. The whole reason the answer is a probability is because we don't have additional information; if we did, it would be a different problem, and if such information exists, it doesn't matter because we don't have it.
The point of the problem is just to show that by being a little tricky about which dog we're actually guessing the gender of, you can end up with an unintuitive (1/3) probability. If you want to add extra factors, go ahead; but if you do, you're solving a different problem. The answer to this problem, as stated, is 1/3.
This problem isn't really as complex as a lot of these posts make it sound. Someone brought up stuff about breeder selection, and that we don't know how he knows one of the dogs is male. But unless we assume something illogical like that some unknown circumstance might make both of them male, it doesn't even matter how he knows this. All that matters is that we know one or more of the dogs is male; where this information came from is irrelevant. There are only 2 steps here:
1. One or more of the dogs is male, so the possibilities are M/M, M/F, and F/M. Presumably, these are all equally probable.
2. The "Other" dog is best described as the dog not chosen from the pair to prove that one of them is male, so in each set the gender of this dog is M, F, and F, respectively. And that's it. 1/3 chance of being male.
The source of the dogs does not matter, and neither does anything else that could affect the dogs' genders, simply because we don't know any of this. It's just as likely that they could only be both male for some reason as it is that one of them must be female for some reason. If we had any information like that, we would already know the answer to 100% accuracy, and this wouldn't even be a problem. The whole reason the answer is a probability is because we don't have additional information; if we did, it would be a different problem, and if such information exists, it doesn't matter because we don't have it.
The point of the problem is just to show that by being a little tricky about which dog we're actually guessing the gender of, you can end up with an unintuitive (1/3) probability. If you want to add extra factors, go ahead; but if you do, you're solving a different problem. The answer to this problem, as stated, is 1/3.
Ah, the problem is that I accidentally made the same error that I've been complaining about(I'm allowed to make mistakes, too). The statements are trying to say something about what the bather actually means when we have no such information explicit or implicit in the problem. The statements pick out a particular puppy, the one the bather is referencing, but we don't really know which one he is referencing. So, we can't really make any reliable guesses along those lines. We don't know how he came by the knowledge that at least one puppy is male such to even allow him to reference a specific one. As has been pointed out in earlier posts, there are many different background scenarios that could lead to the bather's knowledge of the puppies. So, I need to change my example to exclude use of knowledge for which I do not have access, directly or indirectly, or can not know reliably in the context of the problem.Cheeze_Pavilion post=18.73797.842392 said:
So, I propose one has to take a more blind approach to only consider the fact there are two puppies and at least one is male. Nothing else. Although, we do make an assumption in all this(I don't know if you missed this one, too, or not, Cheeeze). We are making an assumption that the puppies are only male or female, not something like hermaphrodite. So, then one could say that we only work with the information that there are two puppies and at least one is male with the assumption that the gender of any single puppy is either male or female. We continue to label the dogs dog1 and dog2 only for the purpose of distinguishing the dogs in our mental construct. Now, starting dog1, we ask what are the possible genders. They are male or female. If dog1 is male, then dog2 may be male or female. If dog1 is female, then dog2 can only be male because of the condition there be at least one male. So, to draw out this tree you mentioned, we have
case 1: dog1 is male
dog1/dog2
M/M
M/F
case 2: dog1 is female
dog1/dog2
F/M
Now, let's give some consideration to the labels "the other one" and "not the other one". I propose these labels generate same tree as above by replacing the dog1 and dog2 labels because these labels also can not refer to a specific dog. We can not say "not the other one" must refer to the known male dog because that picks a specific dog to reference, and we don't know which dog is being referenced as the known male dog. We don't know which dog to pick as "not the other one" to be the target of the bather's answer. In fact, we don't know if the bather is even referencing any dog at all since he could have come by his knowledge through different channels(a possibility that has been shown in other posts above). So, suggesting that "not the other one" always refers to the male dog is making specificity that is not actually given in the problem.
You could say that the labels dog1 and dog2 specify a particular dog. However, the truth is that they do not. I could just as easily swap the labels between the dogs. These labels are not meant to reference the dogs, only to distinguish them.
Do you see anything wrong with what I have done?
As a civil enginering student in industrial mathematics it pains me to se so much discussion over such a simple question.
First of all to Cheeze_Pavilion: We assume that M/M, F/F, M/F and F/M are equally probable because its the logical thing to do. The chance of a single dog becoming male or female is naturally assumed to be equal, since its logical to think that an equal number of male and female dogs exist in the world, as with humans. It's also natural to assume that the gender of one pupppy is independent of the other, since we have no information to calculate an answer if this is not the case. This gives that each of the 4 posibilities listed above are equally probable.
Now the correct answer comes down to interpretation. If we know that at least a spesific dog is male, say dog 1. Then we are left with 2 equally likely options M/F and M/M, and the probability is 50%. If we know that at least one of the dogs is male, but not which one, then we have only eliminated F/F. Thus our desired result is one of 3 equaly likely ones and the probability is 1/3. Can we please move on now! The question isn't hard it's just very porely written.
First of all to Cheeze_Pavilion: We assume that M/M, F/F, M/F and F/M are equally probable because its the logical thing to do. The chance of a single dog becoming male or female is naturally assumed to be equal, since its logical to think that an equal number of male and female dogs exist in the world, as with humans. It's also natural to assume that the gender of one pupppy is independent of the other, since we have no information to calculate an answer if this is not the case. This gives that each of the 4 posibilities listed above are equally probable.
Now the correct answer comes down to interpretation. If we know that at least a spesific dog is male, say dog 1. Then we are left with 2 equally likely options M/F and M/M, and the probability is 50%. If we know that at least one of the dogs is male, but not which one, then we have only eliminated F/F. Thus our desired result is one of 3 equaly likely ones and the probability is 1/3. Can we please move on now! The question isn't hard it's just very porely written.
Except this is exactly how probability works. If outcomes can be eliminated as having zero probability, then the resulting probabilities of the remaining outcomes must sum to 1. Do you not agree that this is true? Are you implying that if the FF outcome is removed the other outcomes should maintain probabilities of 25% and 50%, giving a total probability of only 75%? Renormalizing the probability of outcomes is a standard procedure when the number of outcomes changes. You sum the old probabilities, and then divide each by that sum to obtain the new probabilities. The book that Alex_P mentioned even shows this in the decision trees that it uses.Cheeze_Pavilion post=18.73797.844760 said:
The other thing is that I did not eliminate the FF option for no reason. I'm not sure why you are trying to make it sound like I did. I eliminated it because we are told that there is at least one male in the pair of puppies. The FF option is inconsistent with that fact. I can not have both puppies female and still have at least one be male. That's nonsensical, hence, the reason I eliminated it.
Except that's just it, they are not equally probable. The mixed pair is twice as likely as the male-only pair. This is because the mixed pair can be manifest as MF or FM, and swapping the order in any one of those configurations does not return you to the same configuration. So, we have to consider each of these configurations unique and distinct from each other. However, for the MM pair, swapping the the order does return you to the same configuration. So, we can not consider MM and MM to be unique and distinct from each other. So the MM pair only has one unique and distinct configuration that can manifest it, whereas the mixed pair has two unique and distinct configurations that can manifest it.Cheeze_Pavilion post=18.73797.844760 said:
First, if you don't choose a point of view, it's difficult to find a solution to the problem. Second, I chose the point of view of someone who doesn't know which puppy is being referenced because we don't have any such indication. It would be different if we were given a name, a tag, or just something that let us know specifically that that particular puppy is the one that is being designated as the known male puppy. In that case, the answer would indeed be 50%.Cheeze_Pavilion post=18.73797.844760 said:
I do have to comment, Cheeze, that you have this nasty habit of saying that I am doing things I'm not doing. You also have a nasty habit of trying to twist my words to distract from the logic at hand. You also keep accusing me of doing things for no reason or not knowing what I am doing or saying. I admit to making mistakes, but that does not mean I don't have reasoning or understanding.
I have been stating that unreliable or inaccessible information and questions whose answers are unreliable or inaccessible have to be discarded because using such information as a premise does not lead to a reliable conclusion. For this particular problem, I have been saying that "the other one" does not provide reliable information as to which puppy is being referenced because the previous statements do not make a reference to a specific puppy, i.e. the "not the other one". So, I have been contending that any information that tries to attach to a specific puppy can not be used. Further, the conjuration of background processes and other effects which may change the probabilities of the puppies' gender also can not be used because those processes and effects are not directly or indirectly observable from the context of the problem, and no information regarding these processes or effects can be reliably obtained from the context of the problem. An uncountable infinity of such conjurations are possible, and this invites a person to be able to say anything that he wants to say and not be counted incorrect.
Now, let me ask you this, and with these questions I'll shut up and never bother you again on this: Do you have an example that contradicts the idea that an unreliable premise does not create reliable conclusions? Do you have at least one example in which an unreliable premise does lead to a reliable conclusion(and coming up with an example of finding the right answer from the wrong premises is not an example of finding a reliable conclusion from unreliable premises)? Do you think that it is reliable premise that "the other one" specifies a particular puppy? If so, why do you think it is a reliable premise?
Because there are no other reasonable presumptions, unless you know something about biology that no one else does.Cheeze_Pavilion post=18.73797.844772 said:
No, really. Let's say, for some reason we don't know, both of the dogs must be male. Fine. It's also possible that one has to be female, also for a reason we don't know. It doesn't matter, because we don't know. All other things being equal--an assumption we have to make, since we don't know anything else--two randomly chosen dogs each have a 50% chance of being male and a 50% chance of being female (+ or - 1% or so). So, M/M, F/M, M/F, and F/F are all equally likely. But F/F is impossible, and...well, I think I've explained the rest enough times by now.
This is not as complex as you think. All we can use to determine the dogs' genders is biology and the answer to the shopkeeper's question, and from only that information, the possibilities are all equally likely. For them to not be equally likely, we'd have to have some other piece of information. But we don't. So they are.
Sorry, but this is how you'd have to set it up:Cheeze_Pavilion post=18.73797.844760 said:
25 % Male (pair)
25 % Female (pair)
50 % (Mixed) pair
and then this:
33 % Male (pair)
Female (pair) (0 %)
66 % (Mixed) pair
Ruling out the female female option rules out 25 percent of your original probability space, leaving 75 percent behind. So, redistribute that 75 percent across the entire probability space and you get:
MM = 25/75 = 1/3 = 33 percent
Mixed = 50/75 = 2/3 = 66 percent
Well if you throw out biology, the answer to the problem is "who knows?", because you have no possible way of obtaining any probabilities at all to solve the problem. You can't just use the shopkeeper's words, because that's like assuming the dogs are aliens or robots or something so that they have no connection to biology. Surely we can assume the dogs are dogs?Cheeze_Pavilion post=18.73797.845373 said:
No, it doesn't. There is only 1 way to arrange 2 male puppies (M/M), but there are 2 ways to arrange a female puppy and a male puppy (M/F and F/M), if you arrange them into sets called "this is puppy A" and "this is puppy B". This is actually an example of entropy, and it's really not something you can argue with.Cheeze_Pavilion post=18.73797.845448 said:
Can we just move on to something else? This is getting ridiculous.
No, it means the combination is twice as likely, if there are two permutations of it. You're the one who misunderstands the meaning of these two probability terms. There are two permutations of the male female pair here, therefore that combination is twice as likely, while each permutation is equally likely.Cheeze_Pavilion post=18.73797.845512 said:
To use an old table top gaming joke, I'm sorry, I'm afraid I have 0 ranks in Knowledge (What the Hell You're Talking About)Cheeze_Pavilion post=18.73797.845521 said:
I'm afraid that you're now starting to convince me that you know even less about probability than I would have guessed. A combination, like that of a male female pair, has a probability defined by the number of permutations it contains. For instance, out of ten flipped coins, there are ten permutations of the combination, 1 heads and 9 tails. Each permutation is equally likely. There are 2 to the tenth possible permutations of this situation, and so this combination's probability is equal to 10/(2^10).
Sure, it can be like that. But it isn't. The shopkeeper didn't say "They were picked from a set of pairs of dogs, which were [describe each set], and I don't know which one got picked." All (she?) said was that the genders of the dogs are not known, and so they could be a pair or they could not be a pair.Cheeze_Pavilion post=18.73797.845521 said:
This is...absurd. I give up, this argument is hopeless.Cheeze_Pavilion post=18.73797.845521 said: